* Luca Villani (villo@xxxxxxxxxxxxxxxx) [100208 15:31]:
>
> Lo sapete che tomcat quando risolve un indirizzo poi se ne frega del
> TTL, lo scrive su una tavoletta di argilla, la cuoce e la tiene
> direttamente negli archivi reali di Babilonia, vero?
>
Villo, sei entrato di diritto nei miei fortunes
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.*. finelli
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Let {(X_1, X_2)} be drawn uniformly at random from the set {\{ (-1,-1),
(-1,+1), (+1,-1), (+1,+1) \}}. Then the random variables {X_1}, {X_2},
and {-X_1} all individually have the same distribution, namely the
signed Bernoulli distribution. However the pairs {(X_1,X_2)},
{(X_1,X_1)}, and {(X_1,-X_1)} all have different joint distributions:
the first pair, by definition, is uniformly distributed in {\{ (-1,-1),
(-1,+1), (+1,-1), (+1,+1) \}}, while the second pair is uniformly
distributed in {\{(-1,-1),(+1,+1)\}}, and the third pair is uniformly
distributed in {\{ (-1,+1), (+1,-1)\}}. Thus, for instance, if one is
told that {X, Y} are two random variables with the Bernoulli
distribution, and asked to compute the probability that {X=Y}, there is
insufficient information to solve the problem; if {(X,Y)} were
distributed as {(X_1,X_2)}, then the probability would be {1/2}, while
if {(X,Y)} were distributed as {(X_1,X_1)}, the probability would be
{1}, and if {(X,Y)} were distributed as {(X_1,-X_1)}, the probability
would be {0}. Thus one sees that one needs the joint distribution, and
not just the individual distributions, to obtain a unique answer to the
question.
http://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/
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